Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 98


$$-\frac{1}{4}(\ln x)^{-4}+c.$$

Work Step by Step

Let $u=\ln x$, then $du=\frac{dx}{x}$. Now, we have $$\int \frac{dx}{x(\ln x)^5}=\int \frac{du}{u^5}=-\frac{1}{4}u^{-4}+c\\ =-\frac{1}{4}(\ln x)^{-4}+c.$$
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